Question

Решить уравнение.....

Answer 1

Ответ:

решений нет

Объяснение:

фоточка тебе в помощь ;)

Answer 2

$\displaystyle \tt \frac{1+2x}{6x^2-3x}-\frac{2x-1}{14x^2+7x}=\frac{8}{12x^2-3}\:\:\:\:\: | \: x\ne\frac{1}{2}, \: x\ne-\frac{1}{2}\\\\\displaystyle \tt \frac{1+2x}{6x^2-3x}-\frac{2x-1}{14x^2+7x}-\frac{8}{12x^2-3}=0\\\\\displaystyle \tt \frac{1+2x}{3x(2x-1)}-\frac{2x-1}{7x(2x+1)}-\frac{8}{3(4x^2-1)}=0\\\\ \displaystyle \tt \frac{1+2x}{3x(2x-1)}-\frac{2x-1}{7x(2x+1)}-\frac{8}{3(2x-1)(2x+1)}=0\\\\ \displaystyle \tt \frac{7(2x+1)^2-3(2x-1)^2-56x}{21x(2x-1)(2x+1)}=0$

$\displaystyle \tt \frac{7(4x^2+4x+1)-3(4x^2-4x+1)-56x}{21x(2x-1)(2x+1)}=0\\\\\displaystyle \tt \frac{28x^2+28x+7-12x^2+12x-3-56x}{21x(2x-1)(2x+1)}=0\\\\\displaystyle \tt \frac{16x^2-16x+4}{21x(2x-1)(2x+1)}=0\\\\\displaystyle \tt \frac{4(4x^2-4x+1)}{21x(2x-1)(2x+1)}=0\\\\\displaystyle \tt \frac{4(2x-1)^2}{21x(2x-1)(2x+1)}=0\\\\\displaystyle \tt \frac{4(2x-1)}{21x(2x+1)}=0\\\\\displaystyle \tt 4(2x-1)=0\\ \displaystyle \tt 8x-4=0\\ \displaystyle \tt 8x=4\\ \displaystyle \tt x=4\div8$

$\displaystyle \tt x=\frac{4}{8}$

$\displaystyle \tt x=\frac{1}{2}$ - не подходит по ОДЗ

Ответ: нет решений